Proposed Problem

Click the figure below to see the complete problem 412 about Triangle, Angles, Congruence.

See also:

Complete Problem 412

Level: High School, SAT Prep, College geometry

## Sunday, January 3, 2010

### Problem 412: Triangle, Angles, Congruence

Labels:
angle,
congruence,
triangle

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Another solution with full detail:

ReplyDeletehttp://img9.imageshack.us/img9/8220/problem412.png

From C draw CE//AB and CE=AB ( see attached picture)

ABEC is a parallelogram and ∠ (ECy)= 2x

Triangle DEC is isosceles => ∠ (DEC)=1/2∠ (ECy)=x

Quadrilateral BDCE is cyclic with 2 opposite sides paralleled

So BDCE is an isosceles trapezoid

And ∠ (BDA)= ∠ (ECy)=2x => Triangles ABD is isosceles

In triangles ABD 4x=180-40 => x=35

My solution:

ReplyDeleteDraw circle 0 (circumcircle of triangle ABD)

Let E be the intersection of circle 0 with BC.

Construct AE and DE

Triangle BED is isosceles since ∠EBD =∠EAD = x

So ∠BAE = ∠BDE = x

Now ∠BDC =40º+2x so ∠EDC=40º+x since ∠BDE= x

Triangles ABE and EDC are similar by SAS

So ∠ DCE= x , finally x=35º since x+x+40º+2x=180º => 4x=140º => x=35º

http://i.imgur.com/K0x5Bvw.png

DeleteLet the angle bisector of A meet BC at E and let BE = p

ReplyDeleteNow < DAE = < DBE = x so ABED is cyclic

So (a-p)a = bc......(1)

Angie bisector theorem

p/(a-p) = c/b .......(2)

(1) X (2) gives us

ap = c^2 hence BC is tangential to Tr ACE at A and so < C = x

Now we know all the angles of Tr. ABC in terms of x and adding 4x+ 40 = 180 and x = 35

Sumith Peiris

Moratuwa

Sri Lanka

Or far more simply because ABED is cyclic BE = DE and Tr.s ABE and CDE are congruent SAS and so < C = x and the result follows

ReplyDeleteProblem 412

ReplyDeleteBrings BE=//DC then BECD is parallelogram so <BcE=x=< BAE=<BEA.Then BECA isisosceles trapezoid so <ACB=<AEB=x . Then 4x+40=180 so x=35.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

It is special case of x = 45-a/4 where a = 40.

ReplyDeleteThe solution is same as Sumith Peiris.

See the

ReplyDeletedrawingDefine point E in BC such as ∠DAE=x and ∠BAE=x

∠DAE=x and ∠DBE=x => ADEB are concyclic

∠BAE=x => ∠BDE=x

∠BDE=x and ∠BDE=x =>ΔBED is isosceles in E

∠BDC= ∠ABD+ ∠BAD= 40+2x and ∠BDE=x => ∠EDC=40+x

=> ∠EDC= ∠ABE= 40+x

=> ΔABE is congruent to ΔCDE (SAS)

=> EA=EC

=>ΔAEC is isosceles in E

=> ∠CAE=∠ACE=x

In ΔCDE : 2x+(40+x)+x= π

=>4x= π-40=140

Therefore x=35

In triangle ABD

ReplyDeletesin(140-2x)/AB=sin2x/BD

AB/BD=sin(140-2x)/sin2x-------(1)

In triangle BCD

sinx/CD=sin(140-3x)/BD

CD/BD=sinx/sin(140-3x)-------(2)

As AB=CD, (1)=(2)

sin(140-2x)/sin2x=sinx/sin(140-3x)

sinxsin2x=sin(140-2x)sin(140-3x)

cosx-cos3x=cosx-cos(280-5x)

cos3x=cos(280-5x)

x=35